Question

A turn of radius $20\text{m}$ is banked for vehicles going at a speed of $36\text{km/h}$. If the coefficient of friction between the road and the tyres is $0.4$, what are the possible speeds of a vehicle so that it neither slips down nor skids up ?

• A. $v_{min}=4\text{m/s};v_{max}=10\text{m/s}$
• B. $v_{min}=8\text{m/s};v_{max}=12\text{m/s}$
• C. $v_{min}=6\text{m/s};v_{max}=10\text{m/s}$
• D. $v_{min}=4\text{m/s};v_{max}=15\text{m/s}$

Answer 1

The correct option is D $v_{min}=4\text{m/s};v_{max}=15\text{m/s}$

Angle of banking for design speed is given by
$\tan \theta =\frac{v_0^2}{Rg}$
Given $v_0=36\text{km/h}=10\text{m/s}$ and
$R=20\text{m}$
$\Rightarrow \tan \theta =\frac{v_0^2}{Rg}=\frac{{10}^2}{20\times 10}=0.5$

For speed greater than design speed: The vehicle has a tendency to slide up. Hence, friction will act downward. Since we are looking for maximum speed without skidding, friction will be limiting i.e $f=\mu N$


In vertical direction :
$\sum F_y=N\cos \theta -\mu N\sin \theta =mg-\left(1\right)$
In horizontal direction :
$\sum F_x=\frac{m{v}_{max}^2}{R}$
$\Rightarrow N\sin \theta +\mu N\cos \theta =\frac{m{v}_{max}^2}{R}-\left(2\right)$
Dividing (2) by (1),
$\frac{N\sin \theta +\mu N\cos \theta }{N\cos \theta -\mu N\sin \theta }=\frac{\frac{m{v}_{max}^2}{R}}{mg}$
$\Rightarrow \frac{\sin \theta +\mu \cos \theta }{\cos \theta -\mu \sin \theta }=\frac{v_{max}^2}{Rg}$
$\Rightarrow \frac{\tan \theta +\mu }{1-\mu \tan \theta }=\frac{v_{max}^2}{Rg}=\frac{0.5+0.4}{1-0.4\times 0.5}$
$\Rightarrow v_{max}=15\text{m/s}$

For speed less than design speed : The vehicle has a tendency to slide down. Hence, friction will act upward. Again, friction will be limiting when speed is minimum (for no slipping)


In horizontal direction:
$\sum F_t=\frac{m{v}_{min}^2}{R}$
$\Rightarrow N^{\prime }\sin \theta -\mu N^{\prime }\cos \theta =\frac{m{v}_{min}^2}{R}-\left(3\right)$
In vertical direction:
$N^{\prime }\cos \theta +\mu N^{\prime }\sin \theta =mg-\left(4\right)$
Dividing (4) by (3)
$\frac{\sin \theta -\mu \cos \theta }{\cos \theta +\mu \sin \theta }=\frac{v_{min}^2}{Rg}$
$\Rightarrow \frac{\tan \theta -\mu }{1+\mu \tan \theta }=\frac{0.5-0.4}{1+0.4\times 0.5}=\frac{v_{min}^2}{20\times 10}$

$\Rightarrow v_{min}\approx 4\text{m/s}$